A ball was moving along a parabola and it breaks up into two parts.
How can I say that the centre of mass will follow the same parabola ?
I know that ( for simplicity I am assuming that the ball is made up of just two particles which break up due to internal stresses) so as acceleration of both the particles is same ($g$) so the C.O.M. acceleration is also ( $a \,\ = \,\ g$) …..equation (1).
Now as there is no external force acting in the horizontal direction, so by $$v_{C.O.M.} = \frac{ M_1V_1 + M_2 V_2}{m_1+m_2}$$ momentum is conserved along horizontal, so the centre of mass velocity along the horizontal should be the same (result 2).
But now still I need to show that the centre of mass will follow the same parabola as before the separation of the two particles of the ball ???
Best Answer
First of all, the system has not been subjected to any external force other than gravity. As a result, the ball breaking into 2 parts due to internal stresses will have its center of mass unmoved.
The entire system, in this case, the ball has not experienced any net external force other than gravity. So there is no net external force acting on the C.O.M. of the ball other than gravity.
So the C.O.M. of the ball itself acts as a body in projectile motion. And just as a projectile falling under gravity without experiencing any drag follows a parabolic path, so does the C.O.M. of the ball in this case.
EDIT: As I have mentioned in my previous comment, under the action of gravity, the ball follows a particular parabolic path.
In this case, the ball is given an initial velocity by the thrower i.e. the thrower gives it some kinetic energy and the ball does work as it rises up converting K.E. to P.E.
I hope this is clear. Because this concept is being used when I am saying "entire work done by COM must be due to gravity". Consider the part of the motion where the ball is falling. There only the gravity is acting on the ball. Now gravity being a conservative force, net work in the ball falling down = net work done in the ball rising up. In other words, net work in a conservative field is 0. But if you consider a small portion of the parabolic path, the ball is being displaced due to gravity and so the ball is converting its potential energy into kinetic energy and work to change its displacement.
Now if the C.O.M. has a movement as a result of which the parabolic path it traces is no longer the same as the original parabola, then it will be like:
Now under gravity it should follow, say, path 2. And you say that it shifts to path 1. In that case there is a displacement and the work done in displacing the C.O.M. requires force by an external agent. Now this cannot be gravity, for gravity does not allow for such displacement, it prefers path 2. So there must be some other force. But there isn't one, as mentioned earlier. So such change in parabolic path is not possible.
As for your 2nd comment, the gravity is doing work on the C.O.M. as gravity is a conservative force. If you could take air drag into account, you might well get it clear that the ball does work in changing its displacement for the field then would be non-conservative.